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Let $R$ be a ring. Define a circle composition $\circ$ in $R$ by $$a\circ b=a+b-ab,\quad a,b\in R.$$ (b) If $S$ is the set of all quasi-regular elements in $R$, then $(S,\circ)$ is a group.

(c) Suppose $R$ has unity. Let $(U(R),\cdot)$ denote the group of units (i.e. invertible elements) of $R$. Show that $(U(R),\cdot)\simeq(S,\circ)$.

For (b) the neutral element is $0$ since $a\circ 0 = a + 0-a0 = 0$ and $0\circ a = 0 + a-0a = 0$, then element is invertible for all $a\in S$ must be $a\circ 1 = a + 1-a1 = 1$ and $1◦a = 1 + a-1a = 1$. I have not been able to prove the closure law of $\circ$, could someone help me? For Part (c), let $\phi :U(R)\rightarrow S$. $\phi$ is given by $a \mapsto a+x-ax$ ?

anankElpis
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Nash
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2 Answers2

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Hint $ $ The calculations are much simpler and more intuitive employing the following observation. Any set bijection $\,h\,:\,R'\to R\,$ serves to transport the ring structure of $\,(R,+,*,0,1)\,$ to $\,(R',\oplus,\otimes,0',1') \,$ by simply defining the operations in $\,R'\,$ to make $\,h\,$ be a ring isomorphism, i.e. so that $\,h(a\oplus b) = h(a)+h(b)\, $ and $\,h(a\otimes b) = h(a)h(b),\,$ yielding the transported operations $$\begin{align} &a \oplus b\, :=\ h^{-1}(h(a) + h(b)),\quad 0' := h^{-1}(0)\\ &a \otimes b\, :=\, \color{darkorange}{h^{-1}}(\color{#0a0}{h(a)}\, *\, \color{#c00}{h(b)}),\quad 1' := h^{-1}(1)\end{align}\ \ \ \ $$

In OP $\ h(x) = 1\!-\!x\ $ so $\,\ a\otimes b \,=\, \color{darkorange}{1\:\!-}(\color{#0a0}{1\!-\!a})\:(\color{#c00}{1\!-\!b})\, = \,a+b-ab$

All of the (universal) ring axioms (or any ring identity) immediately transport from $R$ to $R',$ e.g. the associative law transports as below.

$$ \begin{align} h(\color{#c00}{a\oplus(b\oplus c)}) \,&=\, h(a)+h(b\oplus c)\\ &=\, h(a)+(h(b) + h(c))\\ &=\, (h(a) + h(b))+h(c),\,\ {\rm by}\,\ R\ \rm associative\\ &=\, h(a\oplus b)+h(c)\\ &=\, h(\color{#0a0}{(a\oplus b)\oplus c})\\ \underset{h^{-1}}\Longrightarrow \ \ \color{#c00}{a\oplus(b\oplus c)}\ \ &=\, \ \ \ \ \color{#0a0}{(a\oplus b)\oplus c}\ \ \Longrightarrow\ R'\ \rm is\ also\ associtive\end{align}\qquad\ \ $$

See here for further examples of transporting algebraic structure along a bijection.

Bill Dubuque
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b). You are wrong on first line. Since $$ a\circ 0= a+0-a 0=a $$ $$ 0\circ a= 0+a-0a=a $$ $0$ is the identity element of $(R, \circ)$.

To prove associativity, note $$ (a\circ b)\circ c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc $$ $$ a\circ (b\circ c)=a+(b+c-bc)-a(b+c-bc)=a+b+c-ab-ac-bc+abc $$ So associativity holds. Thus $(R, \circ)$ is monoid with $0$ as identity element.

For any $a\in S$, since it is quasi-regular, there is a $x\in S$ that $a\circ x=x\circ a =0$. This proves that any element of $S$ has an inverse in $S$.

For closure of $S$, we need to prove that for $a,b\in S, \:a\circ b\in S$. Let $a,b\in S$. By definition of right quasi-regular, there are $x,y\in S$ that $a\circ x=b\circ y=0$. By associativity of $S$ $$ (a\circ b)\circ (y\circ x)=(a\circ (b\circ y))\circ x=(a\circ 0)\circ x=a\circ x=0 $$ This means $a\circ b$ is right quasi-regular. Likewise $a\circ b$ can be proved left quasi-regular. Hence $\:a\circ b\in S$.

Hence $(S, \circ)$ is group.

Edit: Proof of c) was added after @Bill Dubuque's post.

c). Define map $\varphi:(R,\circ)\to (R,\cdot), \: \varphi(a)=1-a$. For any $a,b\in R$ $$ \varphi(a)\varphi(b)=(1-a)(1-b)=1-(a+b-ab)=\varphi(a\circ b) $$ Clearly it is bijective and so is monoid isomorphism. For any $a\in R$, if $a\circ x=0$, then $\:\varphi(a\circ x)=(1-a)(1-x)=1$. This means that if $a$ is right quasi-regular, then $1-a$ is (left) unit in $(R,\cdot)$. Likewise if $a$ is left quasi-regular, then $1-a$ is (right) unit in $(R,\cdot)$. So unit in $(R,\cdot)$ is isomorphic to quasi-regular element, i.e. $$ U(R,\cdot)\cong S(R,\circ) $$

Eugene Zhang
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